A reversible engine converts one-sixth of the | vedclass

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(D) Initially,the efficiency $\eta = 1 - \frac{T_2}{T_1} = \frac{1}{6} \dots (i)$When the sink temperature is reduced by $62 \ K$ (or $62^{\circ} C$),

Vedclass · Accepted Answer

$99^{\circ} C, 37^{\circ} C$

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(D) Initially,the efficiency $\eta = 1 - \frac{T_2}{T_1} = \frac{1}{6} \dots (i)$When the sink temperature is reduced by $62 \ K$ (or $62^{\circ} C$),the new efficiency $\eta' = 1 - \frac{T_2 - 62}{T_1} = 2\eta = 2 	imes \frac{1}{6} = \frac{1}{3} \dots (ii)$From $(i)$,$\frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6} \implies T_2 = \frac{5}{6}T_1$Substitute into $(ii)$: $1 - \frac{\frac{5}{6}T_1 - 62}{T_1} = \frac{1}{3}$$1 - \frac{5}{6} + \frac{62}{T_1} = \frac{1}{3} \implies \frac{1}{6} + \frac{62}{T_1} = \frac{1}{3}$$\frac{62}{T_1} = \frac{1}{3} - \frac{1}{6} = \frac{1}{6} \implies T_1 = 372 \ K$Converting to Celsius: $T_1 = 372 - 273 = 99^{\circ} C$$T_2 = \frac{5}{6} 	imes 372 = 310 \ K = 310 - 273 = 37^{\circ} C$Thus,the temperatures are $99^{\circ} C$ and $37^{\circ} C$.

$A$ reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by $62^{\circ} C$,the efficiency of the engine is doubled. The temperatures of the source and sink are

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